ADVANCED PROBLEMS AND SOLUTIONS

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1 ADVANCED PROBLEMS AND SOLUTIONS Edited by RAYMOND E.WHITNEY Lock Have State College, Lock Have, Pesylvaia Sed all commuicatios cocerig Advaced Problems ad Solutios to Raymod E. Whitey, Mathematics Departmet, Lock Have State College, Lock Have, Pesylvaia This departmet especially welcomes problems believed to be ew or extedig old results. Proposers should submit solutios or other iformatio that will assist the editor. To facilitate their cosideratio, solutios should be submitted o separate siged sheets withi two moths after publicatio of the problems. H-95 Proposed by Verer Hoggatt, Jr., Sa Jose State College, Sa Jose, Califoria Cosider the array idicated below: (i) Show that the row sums are F, ^ 2. (ii) Show that the risig diagoal sums are the covolutio of ( F 2 - l ^ = 0 a d { ^ 2, 2 ) } ; = 0, the geeralized umbers of Harris ad Styles. H-96 Proposed by J. B. Roberts, Reed College, Portlad, Orego. (a) Let A 0 be the set of itegral parts of the positive itegral multiples of r, where + ^ 5 T = 43

2 44 ADVANCED PROBLEMS AND SOLUTIONS [Oct. ad let A -, m = 0,, 2,, be the set of itegral parts of the umbers r 2 for 6 A. Prove that the collectio of Z uio of the A.. J (b) Geeralize the propositio i (a). of all positive itegers is the disjoit H-97 Proposed by Lawrece Somer, Uiversity of Hi i o is, Urbaa, Illiois. Let { i r } _ be the t-fiboacci sequeces with positive etries satisfyig the r e c u r- sio relatioship: Fid (t) u t Vu. Z i i=l -i (t) u.lim + u oo SOLUTIONS HYPER -TENSION H-85 Proposed by L Carlitz, Duke Uiversity, Durham, North Carolia. Show that ( - 2x) = ( - D " k ( 2 + k k ) ( 2 k k ) ( - v ) - k 2 F l [-k; + k + ; k + ; x] k=0 ^ / \ / where 2Fi[a,b; c; x] deotes the hypergeometric fuctio. Solutio by the Proposer. We start with the idetity E (2r + 3s)! (y - z) r z s _ r!s!(r + 2s)! ^ +.2r+3s+l - y - z r,s=0 y* Now put y = u + v, z = v, so that frv ; ( 2 r + 3s)L u v V = JLJ rlsl(r + 2s)l,- +,2r+3s+l - u - 2v U r. s=0 * v '

3 972] ADVANCED PROBLEMS AND SOLUTIONS 45 The right-had side of (*) is equal to =0 while the left-had side oo oo E (2r + 3s)l r s V / -x\ k f 2r + 3s + k \, ^ ± U rlsl( r + 2s)l V ^ { ~ } { k J ( U + V) r,s=0 k=0 It follows that,,., \"^,,,k v ^. x k / 2 r + 3s + k \ (2r + 3s)l r s < u + 2 v > = L < u + v) L k=0 r+s=-k ( ' ] \ k J rl8l(r + 2s)l U v E, i x - r - s (r + 2s + )l r s,, v-r-s W r'.si(r + 2s)l ( - r - s)l u v ( u + v ) r+s<. k l N-k (u + v) A-k \7* (s + + k)l k-s s E t ( " } ( - k)l L* sl<k - s)l(k + s)i k=0 s=0 u v ' Takig u =, v = -x, we get ( - 2x) = (-D " k EMV-'MI klkl( - k)t k=0 ( " V)_k ^ ^ + k + ; k + ; X] A CONGRUENCE IN ITS PRIME H-86 Proposed by James Desmod, Florida State Uiversity, Tallahassee, Florida. The geeralized Fiboacci sequece is defied by the recurrece relatio u, + u = u ^, - + where is a iteger ad U 0 ad U^ are arbitrary fixed itegers. For a prime p ad itegers, r, s ad t, u show that _, = U ^ x p+r sp+t (mod p) 3 ^,

4 46 ADVANCED PROBLEMS AND SOLUTIONS [Oct. if p = ± (mod 5) ad + r = s + t, ad that if p = ±2 (mod 5) ad - r = s - t. U _,_ = (-l) r + t U ^ (mod p) p+r sp+t ^ Solutio by the froposer. We have from Hoggatt ad Buggies, "A P r i m e r for the Fiboacci Sequece Part III," Fiboacci Quarterly, Vol., No. 3, 963, p. 65, ad by Fermat's theorem, that p F ^ = V* (? V-^ F F P "i = F F P.. + F _, F P = F F - + F _,_ F (mod p) p+r L^d l+r - r - p+r r - p+r i=0 x / for all ad r. From I. D. Ruggles, "Some Fiboacci Results Usig Fiboacci-Type Sequeces," Fiboacci Quarterly, Vol., No. 2, 963, p. 79, we have that v for all i ad j. Therefore, F.^. = F.. F. + F.F l + l j I j- for all ad r. F _,_ = F F, + F ^ F F + F F. F (mod p) p+r r - r+ p r p- ^ P r e s s, Lodo, 954, p. 50, that We have from Hardy ad Wright, Theory of Numbers, Oxford Uiversity F _- = 0 (mod p) ad F = (mod p) if p = ± (mod 5), ad that F +i - (mod p) ad F = (mod p) if p = ±2 (mod 5). Let p = ± (mod 5) ad + r = s + t. The F _,_ = F F + F ^ F E F ^ (mod p) p+r r - r+ r+ v ' for all ad r. Therefore It is easily verified by iductio that F.* = F ^ = F ^ = F _,_ (mod p). sp+t s+t +r p+r *

5 972] ADVANCED PROBLEMS AND SOLUTIONS 47 U = t L F + U F I 0 - for all. Therefore U p + r s U l F p + r + U 0 F p + r - l = U l F s p + t + U 0 F S p + t - l S U sp + t < m o d P> Now, let p = +2 (mod 5) ad - r = s - t. From page 77 of the referece to Ruggles, we have for all i ad j. Therefore F.^. - F.L. = (-l) j + F.. i+j 3 i-3 F ^ = F F - - F, - F + F F E F F - + F ^- F - 2F, - F + F F p+r r - r+ r r - r+ r+ r for all ad r. Thus = F ^ - L F = ( - l ) r + F (mod p) F r+ r ' - r Hece ( - l ) r + t F _,. = ( - l ) r + t ( - l ) t + F. = ( - l ) r + F = F _,_ (mod p). v v sp+t \ / \ / s_t - r p+r U = p + r U l F + P + r U 0 F B P + r - l U l ( - ) r + t F s P + t + V-l)'- *- F sp+t-l B ( " ) r + t ( U l F s P + t + U 0 F s p + t - l > - < - ) r + t u s P + t ( m o d P>" FIBONACCI IS A SQUARE H-87 Proposed by Ira Gessel, Harvard Uiversity, Cambridge, Massachusetts. Problem: Show that a positive iteger is a Fiboacci umber if ad oly if either or is a square. Solutio by the Proposer. Let F 0 = 0, F 4 =, F - = F + F _- be the Fiboacci series ad L 0 = 2, ^ =, L,- = L + L be the Lucas series. It is well kow that r+ r r - () (-I)" + F* r = F r + F r _ (2) L r = F r + + F r _ x Subtractig four times the first from the square of the secod equatio, we have whece L2 _ 4 ( - l ) r - 4F 2 = (F _,, - F - ) 2 = F 2, r r r+ r - r 5F ( - l ) r = L 2. r r

6 48 ADVANCED PROBLEMS AND SOLUTIONS [Oct. Thus if is a Fiboacci umber, either or is a square. I have two proofs of the coverse. First Proof. We use the theorem (Hardy ad Wright, A Itroductio to the Theory of Numbers, p. 53) that if p ad q are itegers, x is a real umber, ad (p/q) - x < l/2q 2, the p/q is a coverget to the cotiued fractio for x, ad that (Hardy ad Wright, p. 48) the covergets to the cotiued fractio for ( + *s/5)/2 i lowest terms are F - / F. r Assume that 5 2 ± 4 - m 2. The sice m ad have the same parity, k = (m + )/2 is a iteger. The substitutig m = 2k - i 5 2 ± 4 = m 2, we get k 2 - k - 2 = ±, so that k ad are relatively prime ad ±l/ 2 = (k/) 2 - (k/) - = [(k/) - (N/5 + l)/2] [(k/) + (\/5 - l)/2]. Thus (k/) - (s/5 + l ) / 2 = / 2 (k/) + (^5 - l)/2. Sice is a Fiboacci umber, we may assume ^ 2. The (2k - ) 2 = m 2 ^ = ( 2-4) ^ 4 2, so 2k - ^ 2, whece k/ 2= 3/2. Thus (k/) + (\/"5 - l)/2 > 2, so by the two theorems quoted above, k/ = F - / F for some r, ad sice both fractios are reduced, = F. Secod Proof. Assume 5 2 ± 4 = m 2. The m = ±4, so m + \/5 m - N/5 _ ±l " 2 * 2 ' ad sice m ad have the same parity, m + \ r 5, m - V5 _ _ ad * are itegers i Q ( N / 5 ), where Q is the ratioals, ad sice their product is ±, they are uits. It is well kow (Hardy ad Wright, p. 22) that the oly itegral uits of Q(\/I>) are of the form ±x~, where x = ( + \/"5 )/2. The we have (m + ^ 5 ) / 2 = x r = \ (x r + y r ) + f L ^ ^ L N/5, 2 L ^ 5 J where y = -/x. Now x. + y = L ad

7 972] ADVANCED PROBLEMS AND SOLUTIONS 49 ( X r - y r ) / ^ 5 = F r (Hardy ad Wright, p. 48). Thus ( m + N/5) = ( L r + ^ 5 F r >, so = F. SUM SERIES H-89 Proposed by L Carlitz, Duke Uiversity, Durham, North Carolia (Corrected). Show that (2r + 3s)l (a - b y ) r b S y r + 2 s ^ rls!(r + 2s)l H _,,2r+3s+l,, 2 r 9 S = 0 ' ( + ay) - ay - by 2 Solutio by the Proposer. Put so that i - ax ar - bx 2 A m=0 m x m = E * ( - ax - bx 2 )(l - y) ' J m,=0 vi x y Replacig y by x" y this becomes m ~ V ^ -z / J G x y ( - ax - bx 2 )(l - x y) m,=0 Hece that part of the expasio of ( - ax - bx 2 )(l - x V ) that is idepedet of x is equal to

8 420 ADVANCED PROBLEMS AND SOLUTIONS [Oct. (*) - ay - by 2 O the other had, sice ( - ax - bx 2 )(l - x~ y) = ( + ay) - x(a - by) - bx 2 - x" y, we have ( - ax - b x ^ U - x - V ) " = [ ^ a - b ^ + b x ^ + x - V l E r,s,t=0 vt, S (r + s + t)l (a - by) b y r+2s-t rlsltl,- ^ x r + s + t + l x ( + a ^ t The part of this sum that is idepedet of x is obtaied by takig t = r + 2s. We get E (2r + 3s)l (a - b y ) V y r 2 s r T.sI(r + 2s)l,.., v2r+3s+i ( + a y ) r,s=0 Sice this is equal to (*), we have proved the stated idetity. IT'S A MOD WORLD H-90 Proposed by H. H. Fers, Victoria, British Columbia. Prove the followig 2 r F = (mod 5) 2 r L = (mod 5), where F ad L are the Fiboacci ad Lucas umbers, respectively, ad r is the least residue of - (mod 4). Solutio by the Proposer. I a upublished paper by the proposer, it is show that ^\- E ( 2k + ) 5 k k=0

9 9 7 2 ] ADVANCED PROBLEMS AND SOLUTIONS 42 Hece Thus IT] k=l 2 F = (mod 5) x ' 5 k. Let - = 4m + r, where 0 < r < 4. The But Hece To prove use 2 F = 2 F = 2 2 F = (mod 5). 24m = (24 } m ^ ( m o d 5 ) ^ 2 r F = (mod 5). 2 r L = (mod 5), M k=0 % ' (which is derived i the same paper) ad proceed as above* JUST SO MANY TWO'S H-92 Proposed by Roald Alter, Uiversity of Ketucky, Lexigto, Ketucky. If prove that c 3+l M+J.. s (%: JH 3=0 c = N s (N odd, >0). Solutio by the Proposer. I the sequece b k = b k - l " 3 b k - 2 ' ( k " l j b i = b 2 = D > it is easy to show that 2 is the highest power of 2 that divides b. if ad oly if k = 3 (mod 6), Also, by derivig the appropriate Biet formula, it follows that

10 422 ADVANCED PROBLEMS AND SOLUTIONS [Oct. Thus ^im-c-^i-»- 2 j=o k ^ The desired result follows by observig * Editorial Note: Please submit solutios for ay of the problem proposals. blood! We eed fresh A GOLDEN SECTION SEARCH PROBLEM REXH.SHUDDE Waval Post Graduate School, Moterey, Califoria After tirig of usig umerous quadratic fuctios as objective fuctios for examples i my mathematical programmig course, I posed the followig problem for myself: Desig a uimodal fuctio over the (0,) iterval which is cocave, has a maximum i the iterior of (0,), ad is ot a quadratic fuctio. The purpose was to demostrate umerically the golde sectio search.* My first thoughts were to add two fuctios which are cocave over the (0,) iterval with the property that oe goes to - o at 0 ad the other goes to - o at. My two iitial choices were log x ad l/(x - ). The golde sectio search starts at the two poits x t = - (/0) ad x 2 = l/(p where 0 = ( + N / 5 ) / 2. After searchig with 8 poits, I oticed that the iterval of ucertaity still cotaied the first search poit so I thought it about time to fid the locatio of the maximum aalytically. I was dumfouded to discover that if I cotiued idefiitely with the search my iterval of ucertaity would still cotai the iitial search poit. * Douglas J. Wilde, Optimum Seekig Methods, Pretice Hall, Ic. (964).

<aj + ck) m (bj + dk) n. E E (-D m+n - j - k h)(i) j=0 k=0 \ / \ /

$<aj + ck) m (bj + dk) n. E E (-D m+n - j - k h)(i) j=0 k=0 \ / \ /$ ADVANCED PROBLEMS AND SOLUTIONS Edited by RAYMOND j=. WHITNEY Loc Have State College Loc Have # Pesylvaia Sed all commuicatios cocerig Advaced Problems ad Solutios to Raymod E. Whitey, Mathematics Departmet,